Find ∏_{2}^{∞} (n^{3} - 1)/(n^{3} + 1).

**Solution**

Answer: 2/3.

If we factor: n^{3} - 1 = (n - 1)(n^{2} + n + 1), n^{3} + 1 = (n + 1)(n^{2} - n + 1), then most of the terms cancel. Take the product up to n = N. Then the numerator is 1·2·3 ... (N - 1)·7·13·21·31 ... (N^{2} + N + 1) and the denominator is 3·4·5 ... (N + 1)·3·7·13·21·31 ... (N^{2} - N + 1). Hence the product up to n = N is 2/(N(N + 1)) (N^{2} + N + 1)/3 = 2/3 (N^{2} + N + 1)/(N(N + 1)), which tends to 2/3 as N tends to infinity.

© John Scholes

jscholes@kalva.demon.co.uk

30 Nov 1999